x+7 is greater than or equal to 2
Three solutions for inequality in Year 9 math include: Graphing: Plotting the inequality on a graph helps visualize the solution set, showing all the points that satisfy the inequality. Substitution: Testing specific values in the inequality can help determine if they satisfy the condition, providing a practical way to find solutions. Algebraic Manipulation: Rearranging the inequality by isolating the variable can simplify the problem and lead directly to the solution set.
The name for two inequalities written as one inequality is a "compound inequality." This format expresses relationships involving two conditions simultaneously, often using "and" or "or" to connect them. For example, the compound inequality (3 < x < 7) combines two inequalities, (3 < x) and (x < 7).
To determine a solution to an inequality, you need to specify the inequality itself. Solutions vary depending on the inequality's form, such as linear (e.g., (x > 3)) or quadratic (e.g., (x^2 < 4)). Once the inequality is provided, you can identify specific numbers that satisfy it. Please provide the inequality for a precise solution.
negative two-thirds
-(3^(-2)) = -0.111111111
x - 3 is not an inequality.
In solving an inequality you generally use the same methods as for solving an equation. The main difference is that when you multiply or divide each side by a negative, you have to switch the direction of the inequality sign. The solution to an equation is often a single value, but the solution to an inequality is usually an infinite set of numbers, such as x>3.
The question cannot be answered since it contains no inequality.
You solve an inequality in the same way as you would solve an equality (equation). The only difference is that if you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign. Thus, if you have -3x < 9 to find x, you need to divide by -3. That is a negative number so -3x/(-3) > 9/(-3) reverse inequality x > -3
-3x + 7 < -3 -3x < 4 x > -(4/3) ■
The name for two inequalities written as one inequality is a "compound inequality." This format expresses relationships involving two conditions simultaneously, often using "and" or "or" to connect them. For example, the compound inequality (3 < x < 7) combines two inequalities, (3 < x) and (x < 7).
To determine a solution to an inequality, you need to specify the inequality itself. Solutions vary depending on the inequality's form, such as linear (e.g., (x > 3)) or quadratic (e.g., (x^2 < 4)). Once the inequality is provided, you can identify specific numbers that satisfy it. Please provide the inequality for a precise solution.
The inequality sign changes direction. So 2<3 Multiply by -2 and you get -4>-6 (similarly with division).
For the same reason you must flip it when you multiply by a negative number. An example should suffice. 2 < 3 If you multiply by -1, without switching the sign, you get: -2 < -3, which is wrong. Actually, -2 > -3. Look at a number line if you are not sure about this - numbers to the left are less than numbers further to the right. Dividing by a negative number is the same as multiplying by the reciprocal, which in this case is also negative. These signs are strictly the "Greater than" and "Less than" signs. The inequality sign is an = with a / stroke through it. If you divide an inequality by -1 it remains an inequality.
Split it into two parts, solve each part as if it had an "=" sign instead of an inequality. Ex: -6<3x+2<12 -6<3x+2; -8<3x; -8/3<x Other side-- 3x+2<12; 3x<10; x<10/3 Then put them back together. -8/3<x<10/3 *Make sure you remember to switch the direction of the inequality symbol if you are dividing by a negative number.*
The easiest way is to "flip" the inequality symbol end divide by the negative number:Example:6 < 3 - 3s6 - 3 < 3 - 3s -33 < -3s Method a) Divide by negative coefficient and flip the inequality symbol3/-3 > -3s/-3-1 > s or s< -13 < -3s Method b) Full algorithm, eliminate -3s by adding 3s on both sides3 +3s < -3s + 3s3 + 3s < 03 - 3 + 3s < 0 -33s < -33s/3 < -3/3s < -1 Looks familiar? So basically if you perform the full algorithm (method b) you can understand why we flip the inequality symbol when we have to eliminate a negative coefficient but it is faster just to flip the symbol (method a)
Whenever you multiply (or divide) both sides an inequality by a negative number. Example: 3 < 5, therefore -3 > -5 -2x < 10 becomes (by multiplying by -1/2): x > 5 x(x+1) > 0 In this type of inequality, you can divide both sides by x, but you have to consider the two cases, that x > 0 and that x < 0.